A compass needle free to turn in a horizontal plane is placed at the centre of a circular coil of 30turns and radius 12cm. The coil is in a vertical plane malting an angle of `45^(@)` with the magnetic meridian when the current in the coil is 0.35A, the needle points west to east. Determine the horizontal component of earth's magnetic field at the location.

Options

A. `3.9xx10^(-7)"tesla"`
B. `3.9xx10^(5)"tesla"`
C. `8.0xx10^(-5)"tesla"`
D. `7.0xx10^(-7)"tesla"`

Explanation

To determine the horizontal component of the Earth's magnetic field at the location where the compass needle is placed, we can follow these steps:

### Step-by-Step Solution

1. **Convert the Radius to Meters**:
Given the radius \( r = 12 \, \text{cm} \).
\[ r = 12 \, \text{cm} = 12 \times 10^{-2} \, \text{m} = 0.12 \, \text{m} \]

2. **Identify Given Values**:
- Number of turns \( n = 30 \)
- Current \( I = 0.35 \, \text{A} \)
- Angle \( \theta = 45^\circ \)

3. **Calculate the Horizontal Component of the Magnetic Field**:
The formula for the magnetic field at the center of a circular coil is given by:
\[ B = \frac{\mu_0}{4\pi} \cdot \frac{2\pi n I}{r} \]
where \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m/A} \).

4. **Substituting Values into the Formula**:
Substitute \( n \), \( I \), and \( r \) into the equation:
\[ B = \frac{4\pi \times 10^{-7}}{4\pi} \cdot \frac{2\pi \cdot 30 \cdot 0.35}{0.12} \]
Simplifying this gives:
\[ B = 10^{-7} \cdot \frac{2 \cdot 30 \cdot 0.35}{0.12} \]

5. **Calculating the Value**:
First calculate the numerator:
\[ 2 \cdot 30 \cdot 0.35 = 21 \]
Now calculate:
\[ B = 10^{-7} \cdot \frac{21}{0.12} = 10^{-7} \cdot 175 = 1.75 \times 10^{-5} \, \text{T} \]

6. **Finding the Horizontal Component**:
Since the coil is at an angle of \( 45^\circ \) to the magnetic meridian, the horizontal component \( B_H \) is given by:
\[ B_H = B \sin(45^\circ) \]
Knowing that \( \sin(45^\circ) = \frac{1}{\sqrt{2}} \):
\[ B_H = 1.75 \times 10^{-5} \cdot \frac{1}{\sqrt{2}} = \frac{1.75 \times 10^{-5}}{1.414} \approx 1.24 \times 10^{-5} \, \text{T} \]

7. **Final Calculation**:
Thus, the horizontal component of the Earth's magnetic field at the location is approximately:
\[ B_H \approx 1.24 \times 10^{-5} \, \text{T} \]

### Final Answer
The horizontal component of the Earth's magnetic field at the location is approximately \( 1.24 \times 10^{-5} \, \text{T} \).